\(\int \frac {e^{\frac {1}{16} (1+8 \log (2 \log (\frac {e^5+e^x}{x}))+16 \log ^2(2 \log (\frac {e^5+e^x}{x})))} (-e^5+e^x (-1+x)+(-4 e^5+e^x (-4+4 x)) \log (2 \log (\frac {e^5+e^x}{x})))}{(2 e^5 x+2 e^x x) \log (\frac {e^5+e^x}{x})} \, dx\) [330]

Optimal result

Integrand size = 118, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=e^{\left (-\frac {1}{4}-\log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )^2} \] Output:

exp((-1/4-ln(2*ln((exp(5)+exp(x))/x)))^2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\sqrt {2} e^{\frac {1}{16}+\log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )} \sqrt {\log \left (\frac {e^5+e^x}{x}\right )} \] Input:

Integrate[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/ 
x]]^2)/16)*(-E^5 + E^x*(-1 + x) + (-4*E^5 + E^x*(-4 + 4*x))*Log[2*Log[(E^5 
 + E^x)/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]
 

Output:

Sqrt[2]*E^(1/16 + Log[2*Log[(E^5 + E^x)/x]]^2)*Sqrt[Log[(E^5 + E^x)/x]]
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(107\) vs. \(2(25)=50\).

Time = 0.95 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.28, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2704, 27, 27, 2726}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (e^x (x-1)+\left (e^x (4 x-4)-4 e^5\right ) \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )-e^5\right ) \exp \left (\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+8 \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )\right )}{\left (2 e^x x+2 e^5 x\right ) \log \left (\frac {e^x+e^5}{x}\right )} \, dx\)

\(\Big \downarrow \) 2704

\(\displaystyle \int \frac {\sqrt {2} e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )} \left (e^x (x-1)+\left (e^x (4 x-4)-4 e^5\right ) \log \left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )-e^5\right )}{\left (2 e^x x+2 e^5 x\right ) \sqrt {\log \left (\frac {e^x+e^5}{x}\right )}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {2} \int -\frac {e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+1\right )} \left (e^x (1-x)+4 \left (e^x (1-x)+e^5\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+e^5\right )}{2 \left (e^x x+e^5 x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+1\right )} \left (e^x (1-x)+4 \left (e^x (1-x)+e^5\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+e^5\right )}{\left (e^x x+e^5 x\right ) \sqrt {\log \left (\frac {e^5+e^x}{x}\right )}}dx}{\sqrt {2}}\)

\(\Big \downarrow \) 2726

\(\displaystyle \frac {\sqrt {2} \left (e^x+e^5\right ) \left (e^x (1-x)+e^5\right ) e^{\frac {1}{16} \left (16 \log ^2\left (2 \log \left (\frac {e^x+e^5}{x}\right )\right )+1\right )} \sqrt {\log \left (\frac {e^x+e^5}{x}\right )}}{\left (\frac {e^x+e^5}{x^2}-\frac {e^x}{x}\right ) x \left (e^x x+e^5 x\right )}\)

Input:

Int[(E^((1 + 8*Log[2*Log[(E^5 + E^x)/x]] + 16*Log[2*Log[(E^5 + E^x)/x]]^2) 
/16)*(-E^5 + E^x*(-1 + x) + (-4*E^5 + E^x*(-4 + 4*x))*Log[2*Log[(E^5 + E^x 
)/x]]))/((2*E^5*x + 2*E^x*x)*Log[(E^5 + E^x)/x]),x]
 

Output:

(Sqrt[2]*E^((1 + 16*Log[2*Log[(E^5 + E^x)/x]]^2)/16)*(E^5 + E^x)*(E^5 + E^ 
x*(1 - x))*Sqrt[Log[(E^5 + E^x)/x]])/(((E^5 + E^x)/x^2 - E^x/x)*x*(E^5*x + 
 E^x*x))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2704

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)* 
z^(a*b*Log[F]), x] /; FreeQ[{F, a, b}, x]
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 39.94 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\({\mathrm e}^{{\ln \left (2 \ln \left (\frac {{\mathrm e}^{5}+{\mathrm e}^{x}}{x}\right )\right )}^{2}+\frac {\ln \left (2 \ln \left (\frac {{\mathrm e}^{5}+{\mathrm e}^{x}}{x}\right )\right )}{2}+\frac {1}{16}}\)	\(34\)
risch	\(\sqrt {-2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )}\, {\mathrm e}^{{\ln \left (-2 \ln \left (x \right )+2 \ln \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )}{x}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+{\mathrm e}^{x}\right )\right )\right )\right )}^{2}+\frac {1}{16}}\)	\(160\)

Input:

int((((-4+4*x)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(-1+x)*exp(x)- 
exp(5))*exp(ln(2*ln((exp(5)+exp(x))/x))^2+1/2*ln(2*ln((exp(5)+exp(x))/x))+ 
1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x,method=_RETURNVERBOS 
E)
 

Output:

exp(ln(2*ln((exp(5)+exp(x))/x))^2+1/2*ln(2*ln((exp(5)+exp(x))/x))+1/16)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=e^{\left (\log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) + \frac {1}{16}\right )} \] Input:

integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) 
*exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ 
exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori 
thm="fricas")
 

Output:

e^(log(2*log((e^5 + e^x)/x))^2 + 1/2*log(2*log((e^5 + e^x)/x)) + 1/16)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\text {Timed out} \] Input:

integrate((((-4+4*x)*exp(x)-4*exp(5))*ln(2*ln((exp(5)+exp(x))/x))+(-1+x)*e 
xp(x)-exp(5))*exp(ln(2*ln((exp(5)+exp(x))/x))**2+1/2*ln(2*ln((exp(5)+exp(x 
))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/ln((exp(5)+exp(x))/x),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\sqrt {2} \sqrt {-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )} e^{\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )\right ) + \log \left (-\log \left (x\right ) + \log \left (e^{5} + e^{x}\right )\right )^{2} + \frac {1}{16}\right )} \] Input:

integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) 
*exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ 
exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori 
thm="maxima")
 

Output:

sqrt(2)*sqrt(-log(x) + log(e^5 + e^x))*e^(log(2)^2 + 2*log(2)*log(-log(x) 
+ log(e^5 + e^x)) + log(-log(x) + log(e^5 + e^x))^2 + 1/16)
 

Giac [F]

\[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=\int { \frac {{\left ({\left (x - 1\right )} e^{x} + 4 \, {\left ({\left (x - 1\right )} e^{x} - e^{5}\right )} \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) - e^{5}\right )} e^{\left (\log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right )^{2} + \frac {1}{2} \, \log \left (2 \, \log \left (\frac {e^{5} + e^{x}}{x}\right )\right ) + \frac {1}{16}\right )}}{2 \, {\left (x e^{5} + x e^{x}\right )} \log \left (\frac {e^{5} + e^{x}}{x}\right )} \,d x } \] Input:

integrate((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x) 
*exp(x)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+ 
exp(x))/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x, algori 
thm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx={\mathrm {e}}^{1/16}\,{\mathrm {e}}^{{\ln \left (2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )\right )}^2}\,\sqrt {2\,\ln \left (\frac {1}{x}\right )+\ln \left ({\left ({\mathrm {e}}^5+{\mathrm {e}}^x\right )}^2\right )} \] Input:

int(-(exp(log(2*log((exp(5) + exp(x))/x))/2 + log(2*log((exp(5) + exp(x))/ 
x))^2 + 1/16)*(exp(5) + log(2*log((exp(5) + exp(x))/x))*(4*exp(5) - exp(x) 
*(4*x - 4)) - exp(x)*(x - 1)))/(log((exp(5) + exp(x))/x)*(2*x*exp(5) + 2*x 
*exp(x))),x)
 

Output:

exp(1/16)*exp(log(2*log(1/x) + log((exp(5) + exp(x))^2))^2)*(2*log(1/x) + 
log((exp(5) + exp(x))^2))^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {1}{16} \left (1+8 \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )+16 \log ^2\left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )} \left (-e^5+e^x (-1+x)+\left (-4 e^5+e^x (-4+4 x)\right ) \log \left (2 \log \left (\frac {e^5+e^x}{x}\right )\right )\right )}{\left (2 e^5 x+2 e^x x\right ) \log \left (\frac {e^5+e^x}{x}\right )} \, dx=e^{{\mathrm {log}\left (2 \,\mathrm {log}\left (\frac {e^{x}+e^{5}}{x}\right )\right )}^{2}+\frac {1}{16}} \sqrt {\mathrm {log}\left (\frac {e^{x}+e^{5}}{x}\right )}\, \sqrt {2} \] Input:

int((((-4+4*x)*exp(x)-4*exp(5))*log(2*log((exp(5)+exp(x))/x))+(-1+x)*exp(x 
)-exp(5))*exp(log(2*log((exp(5)+exp(x))/x))^2+1/2*log(2*log((exp(5)+exp(x) 
)/x))+1/16)/(2*exp(x)*x+2*x*exp(5))/log((exp(5)+exp(x))/x),x)
 

Output:

e**((16*log(2*log((e**x + e**5)/x))**2 + 1)/16)*sqrt(log((e**x + e**5)/x)) 
*sqrt(2)